Question

The coefficient of \( x^2 \) in the expansion of the determinant \( \begin{vmatrix} x^2 & x^3+1 & x^5+2 \\ x^3+3 & x^2+x & x^3+x^4 \\ x+4 & x^3+x^5 & 2^3 \end{vmatrix} \) is:

• A. $-10$
• B. $-8$
• C. $-2$
• D. $-6$
• E. $8$

Hint:

To find the coefficient of $x^n$ in a determinant, you can differentiate the determinant $n$ times using the rule of differentiating rows/columns and evaluate at $x=0$, then divide by $n!$.

Answer 1

The Correct Option is A

Concept: The determinant of a matrix where entries are polynomials is itself a polynomial in $x$. To find the coefficient of $x^2$ in the expansion of $\det(A)$, we can use the Leibniz formula or expand by minors.
• Let $f(x) = \det(A)$. The coefficient of $x^2$ is given by $\frac{f''(0)}{2!}$.
• Alternatively, we can expand the determinant and collect terms that result in a power of $x^2$.

Step 1: Identify the determinant structure from Screenshot 2026-05-01 015359.jpg.
The determinant is: \[ \Delta(x) = \begin{vmatrix} x^2 & x^3+1 & x^5+2 x^3+3 & x^2+x & x^3+x^4 x+4 & x^3+x^5 & 8 \end{vmatrix} \]

Step 2: Expand the determinant specifically looking for $x^2$ terms.
Expanding along the first row ($R_1$): \[ \Delta(x) = x^2 \begin{vmatrix} x^2+x & x^3+x^4 x^3+x^5 & 8 \end{vmatrix} - (x^3+1) \begin{vmatrix} x^3+3 & x^3+x^4 x+4 & 8 \end{vmatrix} + (x^5+2) \begin{vmatrix} x^3+3 & x^2+x x+4 & x^3+x^5 \end{vmatrix} \]

Step 3: Extract coefficients of $x^2$ from each part.

• Term 1: $x^2 [ 8(x^2+x) - (x^3+x^4)(x^3+x^5) ]$.
Since there is an $x^2$ outside, we only need the constant term from the bracket to get an $x^2$ term overall. The bracket's lowest power is $x$, so the constant term is $0$. Contribution $= 0$.
• Term 2: $-(x^3+1) [ 8(x^3+3) - (x+4)(x^3+x^4) ]$.
$[ 8x^3 + 24 - (x^4 + x^5 + 4x^3 + 4x^4) ] = [ -x^5 - 5x^4 + 4x^3 + 24 ]$. Multiplying by $-(x^3+1)$, the only way to get $x^2$ is (constant from $(x^3+1)$) $\times$ ($x^2$ from bracket). There is no $x^2$ in the bracket. Contribution $= 0$.
• Term 3: $(x^5+2) [ (x^3+3)(x^3+x^5) - (x+4)(x^2+x) ]$.
$[ (x^6 + x^8 + 3x^3 + 3x^5) - (x^3 + x^2 + 4x^2 + 4x) ] = [ x^8 + x^6 + 3x^5 + 2x^3 - 5x^2 - 4x ]$. Multiplying by $(x^5+2)$, the $x^2$ term is formed by $2 \times (-5x^2) = -10x^2$.

Step 4: Sum the coefficients.
Total coefficient $= 0 + 0 - 10 = -10$.